Server-Side Request Forgery
Problem
$url = $_GET['url'];
$response = file_get_contents($url); // Ungeprüfte URL
echo $response;Lösung
$allowed_hosts = ['example.com', 'api.example.com'];
$parsed_url = parse_url($_GET['url']);
if (!in_array($parsed_url['host'], $allowed_hosts)) {
die("Ungültige URL.");
}
$response = file_get_contents($_GET['url']);
echo $response;Zuletzt aktualisiert